Question

我正在尝试将我的应用与我的数据库连接。

我制作这个脚本php：

<?php

/*ci colleghiamo al database(attenti perchè se lavorate in locale 
l'host è 10.0.2.2 e non 127.0.0.1)*/
 mysql_connect("localhost","mydb","")
    or die("Impossibile connettersi al server MySQL.\n"); 
//selezioniamo il db a cui ci vogliamo connettere
mysql_select_db("mydb")
    or die("Impossibile aprire il database.\n");
//creo la tabella nel database
mysql_query("CREATE TABLE IF NOT EXISTS `mydb`.`survey` (
                    `Museum_ID` INT NOT NULL,
                `Game_ID` INT NOT NULL,
                `Vote` INT NOT NULL,
                PRIMARY KEY (`Museum_ID`))
                ENGINE = InnoDB;"); 
$idMuseum = $_POST['Museum_ID'];
$idGame = $_POST['Game_ID'];
$Vote = $_POST['Vote']
$result = mysql_query("INSERT INTO `mydb`.`survey` (`Museum_ID`, `Game_ID`, `Vote`) VALUES    ($idMuseum, $idGame, $Vote);");
$row = mysqli_fetch_array($result);
$data = $row[0];
if($data){
echo $data;


mysql_close();
?>

这是我的Java代码：

public class MakePost extends AsyncTask<String, Void, Void>{

     @Override
        protected void onPreExecute() {
            super.onPreExecute();

            //toggleUI(0);

            pDialog = new ProgressDialog(Survey.this);
            pDialog.setMessage("Sending data. Please wait...");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(false);
            pDialog.show();
        }

     protected Void doInBackground(String... args) {

            List<NameValuePair> parametriDaInviare = new ArrayList<NameValuePair>(2);
            JSONObject json = null;
            int success = -1;
            updatesAv = false;


            try {
                if(checkedButton == R.id.lowChoise){
                    //aggiungo alla lista parametri il voto
                    parametriDaInviare.add(new BasicNameValuePair("Vote", "1"));
                    parametriDaInviare.add(new BasicNameValuePair("Museum_ID", "1"));
                    parametriDaInviare.add(new BasicNameValuePair("Game_ID", "1"));

                     // get JSON Object by using POST method
                    json = jParser.makeHttpRequest(url_Survey, "POST", parametriDaInviare);             
                    try
                    {
                        Log.d("PHP Response", json.toString());
                        success = json.getInt(TAG_SUCCESS);
                        if(success == 1)
                        {   
                            popupWindow.dismiss();
                            updatesAv =true;

                        }
                        else {
                            updatesAv = false;
                        }
                        Log.d("UPDATES CHECK", "RESULT: " + updatesAv);


                    }catch (JSONException e) 
                    {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                        updatesAv = false;
                    }
                }
                if(checkedButton == R.id.mediumChoise){
                    //do something

                    parametriDaInviare.add(new BasicNameValuePair("Vote", "2"));
                    parametriDaInviare.add(new BasicNameValuePair("Museum_ID", "1"));
                    parametriDaInviare.add(new BasicNameValuePair("Game_ID", "1"));                     
                     // get JSON Object by using POST method
                    json = jParser.makeHttpRequest(url_Survey, "POST", parametriDaInviare);             
                    try
                    {
                        Log.d("PHP Response", json.toString());
                        success = json.getInt(TAG_SUCCESS);
                        if(success == 1)
                        {   
                            popupWindow.dismiss();
                            updatesAv =true;

                        }
                        else {
                            updatesAv = false;
                        }
                        Log.d("UPDATES CHECK", "RESULT: " + updatesAv);


                    }catch (JSONException e) 
                    {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                        updatesAv = false;
                    }
                }
                if(checkedButton == R.id.highChoise){
                    //do something

                    parametriDaInviare.add(new BasicNameValuePair("Vote", "3"));
                    parametriDaInviare.add(new BasicNameValuePair("Museum_ID", "1"));
                    parametriDaInviare.add(new BasicNameValuePair("Game_ID", "1"));                      // get JSON Object by using POST method
                    json = jParser.makeHttpRequest(url_Survey, "POST", parametriDaInviare);             
                    try
                    {
                        Log.d("PHP Response", json.toString());
                        success = json.getInt(TAG_SUCCESS);
                        if(success == 1)
                        {   
                            popupWindow.dismiss();
                            updatesAv =true;

                        }
                        else {
                            updatesAv = false;
                        }
                        Log.d("UPDATES CHECK", "RESULT: " + updatesAv);


                    }catch (JSONException e) 
                    {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                        updatesAv = false;
                    }
                }



            } catch (Exception e) {
                // TODO Auto-generated catch block
                // da implementare
            }



     return null;    
     }


     protected void onPostExecute(String file_url) {
            // dismiss the dialog after getting all data
            pDialog.dismiss();

        }

当我发送数据时，我有这个错误：

11-03 13:20:55.918: E/JSON Parser(11238): Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject

我对PHP的经验不多，我不知道我的代码中是否有错误。有人能帮我吗？

由于

Answer 1

变化
echo $ data

要 echo json_encode（$ data）;

在此行的末尾添加分号

$ Vote = $ _POST ['投票'];

Answer 2

首先你应该删除$row = mysqli_fetch_array($result); 这是一个无需取的插入！更不用说你在混音 mysqli和mysql扩展
使用mysqli扩展，您将接触到SQL注入
最后echo $data;不是JSON，它会破坏你的应用程序尝试解析json数据。
缺少分号$Vote = $_POST['Vote'];

if(!$result){
    echo json_encode(array('success'=> 0));
}else{
    echo json_encode(array('success'=> 1));
}

Answer 3

您的服务器主要是将HTML传回给应用程序，然后尝试解析它失败。正如错误所述，响应以br开头。请检查您的服务器是否收到了正确的回复

Answer 4

你需要将你的php响应更改为这样 echo Json_encode（$ data）; 因为它正在发送飞机html数据，说明你获得JSONEXCEPTION的原因

Answer 5

您的查询中有一个错误位置的分号：

更改此内容：VALUES ($idMuseum, $idGame, $Vote);");

到 VALUES ($idMuseum, $idGame, $Vote)");

正如amit kumar已经提到过的，你应该使用echo json_encode($data);

Android - 关于httppost的JsonException

5 个答案: